Problem: Evaluate $~~\int \tan^{-1}x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\tan^{-1}x-\ln(x^2+1)+C$ (Choice B) B $x\tan^{-1}x-\ln x+C$ (Choice C) C $\tan^{-1}x-\dfrac12\ln(x^2+1)+C$ (Choice D) D $x\tan^{-1}x-\dfrac12\ln(x^2+1)+C$
Answer: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \tan^{-1}x~$ and $~dv=dx\,$. Then $~du = \dfrac{1}{x^2+1}~$ and $~v = \int dx = x\,$. Integration by parts gives $ \int \tan^{-1}x\,dx = x\tan^{-1}x-\int\dfrac{x}{x^2+1}\,dx$ $ ~~~~~\,=x\tan^{-1}x-\dfrac12\ln(x^2+1)+C\,$.